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Difference between a spanning and linear combination?

  1. Aug 15, 2010 #1
    what is the difference of a span of a vector and a linear combination of a vector?
     
  2. jcsd
  3. Aug 15, 2010 #2

    quasar987

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    Given a set of vectors v_1,...,v_n, a linear combination of those vectors is a vector of the form a_1v_1+...+a_nv_n, for a_1,...,a_n some scalars. The span of the vectors v_1,...,v_n is the set of all the linear combinations of those vectors. I.e., it is the set whose elements are the vectors of the form a_1v_1+...+a_nv_n for all possible values of the coefficients a_1,...,a_n.
     
  4. Aug 15, 2010 #3
    hmm, sorry for interrupting, but how to know when the span of some vector are equal?
    ie: The span of the vectors v_1,...,v_n = The span of the vectors u_1,...,u_k.
     
  5. Aug 15, 2010 #4

    quasar987

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    One way to do it would be to check that every vector u_i is itself a linear combination of the v_j, and conversely, that every vector v_j can be written as a linear combination of the u_i.

    If this is so, then span(u_1,...,u_k)=span(v_1,...,v_n). If not, then the spans are not equal.

    Make sure you see why.
     
  6. Aug 15, 2010 #5
    i can't clearly see it, every element in span(u_1,...,u_k) is then the element of span(v_1,...,v_n), and conversely,

    then span(u_1,...,u_k) is subset of span(v_1,...,v_n), and conversely

    is that really correct?
     
  7. Aug 15, 2010 #6

    quasar987

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    Suppose that for each i=1,...,k,

    [tex]u_i = \sum_{j=1}^nc_i^jv_j[/tex]

    Then, for an arbitrary linear combination of the u_i,

    [tex]\sum_{i=1}^ka^iu_i=\sum_{i=1}^ka^i\left(\sum_{j=1}^nc_i^jv_j\right)=\sum_{j=1}^n\left(\sum_{i=1}^ka^ic_i^j\right)v_j[/tex]

    (a linear combination of the v_j !) This shows that [itex]\mathrm{span}(u_1,\ldots,u_k)\subset \mathrm{span}(v_1,\ldots,v_n)[/itex].

    And in the same way, if each v_j can be written as a linear combination of the u_i, we obtain [itex]\mathrm{span}(v_1,\ldots,v_n)\subset \mathrm{span}(u_1,\ldots,u_k)[/itex].

    And so in that case, [itex]\mathrm{span}(v_1,\ldots,v_n)= \mathrm{span}(u_1,\ldots,u_k)[/itex].

    On the other hand, if for instance, u_i cannot be written as a linear combination of the v_j's, then [itex]\mathrm{span}(v_1,\ldots,v_n)\neq \mathrm{span}(u_1,\ldots,u_k)[/itex] since [itex]u_i\in \mathrm{span}(u_1,\ldots,u_k)[/itex] but [itex]u_i \notin\mathrm{span}(v_1,\ldots,v_n)[/itex].
     
  8. Aug 15, 2010 #7
    thanksssssssssss, i can see it now
     
  9. Aug 15, 2010 #8

    HallsofIvy

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    And please note that these are the span of a set of vectors and the linear combination of a set of vectors. Typically the set contains more than just one vector!

    A linear combination is single sum of scalars times vectors in the set. The span is the collection of all possible linear combinations of the set.
     
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